## Clock Frequency

Question 1 |

The floating point unit of a processor using a design D takes 2t cycles compared to t cycles taken by the fixed point unit. There are two more design suggestions D1 and D2. D1 uses 30% more cycles for fixed point unit but 30% less cycles for floating point unit as compared to design D. D2 uses 40% less cycles for fixed point unit but 10% more cycles for floating point unit as compared to design D. For a given program which has 80% fixed point operations and 20% floating point operations, which of the following ordering reflects the relative performances of three designs? (Di > Dj denotes that Di is faster than Dj)

D1 > D > D2 | |

D2 > D > D1 | |

D > D2 > D1 | |

D > D1 > D2 |

Question 1 Explanation:

T = 0.8 × time taken in fixed point + 0.2 × time taken in floating point

D = 0.8 × t + 0.2 × 2t = 1.2t

D1 = 0.8 × 1.3t + 0.2 × 0.7 × 2t = 1.32t

D2 = 0.8 × 0.6t + 0.2 × 1.1 × 2t = 0.92t

⇒ D2 > D > D1

D = 0.8 × t + 0.2 × 2t = 1.2t

D1 = 0.8 × 1.3t + 0.2 × 0.7 × 2t = 1.32t

D2 = 0.8 × 0.6t + 0.2 × 1.1 × 2t = 0.92t

⇒ D2 > D > D1

There is 1 question to complete.